Question: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 + 42}{x + 9} = \dfrac{-2x + 105}{x + 9}$
Multiply both sides by $x + 9$ $ \dfrac{x^2 + 42}{x + 9} (x + 9) = \dfrac{-2x + 105}{x + 9} (x + 9)$ $ x^2 + 42 = -2x + 105$ Subtract $-2x + 105$ from both sides: $ x^2 + 42 - (-2x + 105) = -2x + 105 - (-2x + 105)$ $ x^2 + 42 + 2x - 105 = 0$ $ x^2 - 63 + 2x = 0$ Factor the expression: $ (x + 9)(x - 7) = 0$ Therefore $x = -9$ or $x = 7$ However, the original expression is undefined when $x = -9$. Therefore, the only solution is $x = 7$.